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Homework Help: Double integral evaluation

  1. Apr 9, 2005 #1
    More fun yaaay
    evaluate [tex] \int \int \int_{G} x^2 yz dx dy dz [/tex]
    where G is bounded by plane z=0, z=x, y=1, y=x

    certrainly zi s bounded below by 0 and above by x. and y is boundedbelow by 1 and above by x. having a hard time picturing this...

    i dont think this would pictured how the double integrals were, is there a way to visualize this?? Let x come out of the apge, y go right and z upwards
    then z=0 is the x y plane, y =1 come out the page y=x come out of the and goes right and finally z=x come out the page and goes upward
    but i dont think this helped determining the bounds did it??
    Plase help
     
  2. jcsd
  3. Apr 9, 2005 #2

    xanthym

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    SOLUTION HINTS:
    Try the following:

    [tex] 1: \ \ \ \ \int_{x=0}^{1} \int_{y=1}^{x} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dy \, dx \ \ = [/tex]

    [tex] 2: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=1}^{x} y \ \left( \int_{z=0}^{x} z \, dz \right) \ dy \right ) \ dx \ \ = [/tex]

    [tex] 3: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=1}^{x} y \ \left( \, \left[ z^{2}/2 \right]_{0}^{x} \, \right) \ dy \right ) \ dx \ \ = [/tex]

    [tex] 4: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=1}^{x} y \cdot (x^{2}/2) \ dy \right ) \ dx \ \ = [/tex]

    [tex] 5: \ \ \ \ = \ \ \int_{x=0}^{1} \ (x^{4}/2) \ \left ( \int_{y=1}^{x} y \ dy \right ) \ dx \ \ = [/tex]

    Continue this process until all integrations have been performed and evaluated.


    ~~
     
    Last edited: Apr 9, 2005
  4. Apr 11, 2005 #3

    ehild

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    Homework Helper

    Xanthym, the result is negative, that should not be. Change over the bound of integration with respect to y.

    Stunner, you should sketch a diagram to find the bounds of integration. You have to find a closed volume as the domain of integration. That is a tetrahedron in this case, with vertexes (0,0,0), (1,1,1), (0,1,0) and (1,1,0) Look at the picture. So it is either

    [tex] \int_{x=0}^{1} \int_{y=x}^{1} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dy \, dx [/tex]

    or

    [tex] \int_{y=0}^{1} \int_{x=0}^{y} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dx \, dy [/tex].

    ehild
     
    Last edited: Jun 29, 2010
  5. Apr 11, 2005 #4

    xanthym

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    Science Advisor

    (Bounds on y-integration corrected, now {y="x→1"} instead of previous {y="1→x"}, thanks to ehild's comment.)
    SOLUTION HINTS:

    [tex] 1: \ \ \ \ \int_{x=0}^{1} \int_{y=x}^{1} \int_{z=0}^{x} x^2 \cdot y \cdot z \ dz \, dy \, dx \ \ = [/tex]

    [tex] 2: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=x}^{1} y \ \left( \int_{z=0}^{x} z \, dz \right) \ dy \right ) \ dx \ \ = [/tex]

    [tex] 3: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=x}^{1} y \ \left( \, \left[ z^{2}/2 \right]_{0}^{x} \, \right) \ dy \right ) \ dx \ \ = [/tex]

    [tex] 4: \ \ \ \ = \ \ \int_{x=0}^{1} \ x^2 \ \left ( \int_{y=x}^{1} y \cdot (x^{2}/2) \ dy \right ) \ dx \ \ = [/tex]

    [tex] 5: \ \ \ \ = \ \ \int_{x=0}^{1} \ (x^{4}/2) \ \left ( \int_{y=x}^{1} y \ dy \right ) \ dx \ \ = [/tex]

    [tex] 6: \ \ \ \ = \ \ \int_{x=0}^{1} \ (x^{4}/2) \ \left( \, \left[ y^{2}/2 \right]_{x}^{1} \, \right) \ dx \ \ = [/tex]

    Continue this process until all integrations have been performed and evaluated.


    ~~
     
    Last edited: Apr 11, 2005
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